B

连接$BE$，设$\widehat {DE}$的度数为$\theta$，
则$\angle EBD=\frac{1}{2}θ$，
$\because AE$为直径，
$\therefore \angle ABE=90^{\circ}$，
$\because \angle A=\alpha$，
$\therefore \angle AEB=90-\alpha$，
$\because \angle C=\beta$，$\angle AEB=\angle C+\angle EBC=\beta +\frac{1}{2}θ$，
$\therefore 90^{\circ}-\alpha =\beta +\frac{1}{2}θ$，
解得：$\theta =180^{\circ}-2\left(\alpha +\beta \right)$，
即$\widehat {DE}$的度数为$180^{\circ}-2\left(\alpha +\beta \right)$，
$A$、当$\alpha +\beta =70^{\circ}$时，$\widehat {DE}$的度数是$180^{\circ}-140^{\circ}=40^{\circ}$，故本选项错误；
$B$、当$\alpha +\beta =70^{\circ}$时，$\widehat {DE}$的度数是$180^{\circ}-140^{\circ}=40^{\circ}$，故本选项正确；
$C$、当$\alpha -\beta =70^{\circ}$时，即$\alpha =70^{\circ}+\beta$，$\widehat {DE}$的度数是$180^{\circ}-2\left(70^{\circ}+\beta +\beta \right)=40^{\circ}-4\beta$，故本选项错误；
$D$、当$\alpha -\beta =70^{\circ}$时，即$\alpha =70^{\circ}+\beta$，$\widehat {DE}$的度数是$40^{\circ}-4\beta$，故本选项错误；
故选：$B$.